Question

How many grams of Na2CO3 should be mixed with 5.00 g of NaHCO3 to produce 100 mL of buffer with a pH of 10.00? What volume of 0.01 M NaOH must be added to this solution to achieve a pH of 11.25? (pKa1 is 6.35; pKa2 is 10.53)

Answer 1

Moles of NaHCO₃ = ( mass/molar mass)

= ( 5.00 g / 84.0 g/mol)

= 0.0595 mol .

In the given buffer solution NaHCO₃ act as a weak acid , hence pKa₂ (10.53) is used.

Henderson-Hasselbalch equation is

pH = pKa + log ( ₎

Or, 10.00 = 10.53 + log ( )

Or, log   = - 0.53

or,   = 10^-(0.53)

or, Moles of Na₂CO₃ = 0.295 × 0.0595 = 0.0176

Mass of Na₂CO₃ = moles × molar mass

= 0.0176 mol × 106 g/mol

= 1.86 g.

now,

when pH = 11.25

11.25 = 10.53 + log

or, log ( ) = 0.72

or, = 10^0.72

or, = 5.25

Let x moles of NaOH is added

NaHCO₃ + NaOH Na₂CO₃ + H₂O

	NaHCO3	NaOH	Na2CO3
Before	0.0595	x	0.0176
Change	-x	-x	+x
After	(0.0595-x)	0	(0.0520+x)

Now

= 5.25

Or, 0.0176 + x = (0.0595 - x) × 5.25

Or, 0.0176 + x = 0.312 - 5.25x

Or, 6.25 x = (0.312 - 0.0176)

Or, x = 0.047

Now, moles of NaOH added = 0.047

Volume of NaOH

= (Moles /molarity)

= (0.047/0.01)

= 4.7 L.

Therefore, 4.7 L of NaOH is needed to increase the pH from 10.00 to 11.25 .